Problem: $\text D = \left[\begin{array}{rr}4 & -1 \\ 2 & -1\end{array}\right]$ and $\text A = \left[\begin{array}{rrr}3 & 1 & 0 \\ 2 & 1 & -2\end{array}\right]$ Let $\text {H = DA}$. Find $\text H$. $ {H = }$
Solution: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{D}$ and the first column of $\text{A}$. $ \text {H}=\left[\begin{array}{rr}{4} & {-1} \\ 2 & -1\end{array}\right]\left[\begin{array}{rr} {3} & 1 & 0 \\ {2} & 1 & -2\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(4,-1)\cdot(3,2)\\\\ &=4 \cdot 3 -1\cdot 2\\\\ &=10 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $2 \cdot 1 -1\cdot 1 = 1$ (Choice B) B $4 \cdot 1 -1\cdot 1 = 3$ (Choice C) C $2 \cdot 3 -1\cdot 2 = 4$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rrr}10 & 3 & 2 \\ 4 & 1 & 2\end{array}\right]$